If a circle passes through the points (1,−6),(2,1) and (5,2), then the equation of the circle is
Let the equation of the required circle be
x2+y2+2gx+2fy+c=0 ⋯(i)
According to the problem, the above equation passes through the points (1,−6),(2,1) and (5,2).
Therefore, substituting the coordinates of three points (1,−6),(2,1) and (5,2) successively in equation (i), we get
For the point (1,−6):
1+36+2g−12f+c=0
⇒2g−12f+c=−37 ⋯(ii)
For the point (2,1):
4+1+4g+2f+c=0
⇒4g+2f+c=−5 ⋯(iii)
For the point (5,2):
25+4+10g+4f+c=0
⇒10g+4f+c=−29 ⋯(iv)
Subtracting (ii) from (iii), we get
2g+14f=32
⇒g+7f=16 ⋯(v)
Again, subtracting (ii) from (iv), we get
8g+16f=8
⇒g+2f=1 ⋯(vi)
Now, solving equations (v) and (vi), we get
g=−5 and f=3.
Putting the values of g and f in (iii), we get c=9.
Therefore, the equation of the required circle is
x2+y2−10x+6y+9=0